for $s > 1$, $r\in \mathbb{N}$ and where $d(n)=\#\text{ of divisors of } n$. I've been able to show that $d(n)$, $d(n)^r$ are multiplicative and so my series is Dirichlet. As such, I've broken this into Euler sums and thus transformed the series to

First, your infinite series in Euler product is wrong. It should be $\sum_n \frac{(n+1)^r}{p^{ns}}$.
–
user27126Nov 14 '12 at 1:07

1

One way to proceed is to take log and bound the log of each term in the Euler product. Try to show that each series $\sum_n \frac{(n+1)^r}{p^{ns}}$ is really a sum of finitely many terms (depending on $r$) of constants times $\frac{1}{(log p)^k(1-p^{-s})^k}$, where $k$ goes up to $r$. This allows you to bound log of the series, by sum of log of $\frac{1}{(log p)^k(1-p^{-s})^k}$, and it's easy to see that the latter sum converges for $s > 1$.
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user27126Nov 14 '12 at 1:09

1 Answer
1

It is known (see for instance Apostol's book Introduction to analytic number theory) that for all $\epsilon>0$
$$
d(n)=o(n^\epsilon).
$$
Given $s>1$ choose $\epsilon=\dfrac{s-1}{2\,r}>0$. Then
$$
\sum_{n=1}^\infty\frac{d(n)^r}{n^s}\le C\sum_{n=1}^\infty\frac{n^{r\epsilon}}{n^s}=C\sum_{n=1}^\infty\frac{1}{n^{s-r\epsilon}}=C\sum_{n=1}^\infty\frac{1}{n^{(s+1)/2}}<\infty,
$$
where $C$ is a constant depending on $s$ and $r$.